What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass 256 g ) and the decrease in freezing point is 0.40 K? [2025]

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Solution

Q.
What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass 256 g ${mol}^{- 1}$ ) and the decrease in freezing point is 0.40 K [2025]

1 5.12 K kg ${mol}^{- 1}$

2 3.72 K kg ${mol}^{- 1}$

3 4.43 K kg ${mol}^{- 1}$

4 1.86 K kg ${mol}^{- 1}$

Ans.
(1)

$Δ T_{f} = m k_{f}$

$Δ T_{f} = \frac{w_{solute} (in g) \times 1000}{M_{solute} W_{solvent} (in g)} k_{f}$

$0.4 = \frac{1 \times 1000}{256 \times 50} k_{f}$

$k_{f} = \frac{0.4 \times 256 \times 50}{1000} {K kg mol}^{- 1}$

$= 5.12 {K kg mol}^{- 1}$

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